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NCERT Questions for Class 12 Physics Chapter 5 – Magnetism and Matter
In Class 12 Physics, Chapter 5 covers Magnetism and Matter, a chapter on magnets and the interaction between the magnetic fields. The concepts in this chapter form the basis by which students understand the basics of electromagnetism for applications well beyond the classroom. In this guide, we will discuss some important questions and answers of Chapter 5 Magnetism and Matter.
Important Questions with Solutions of Class 12 Physics Chapter 5 – Magnetism and Matter
1) In the ground state of a hydrogen atom, one electron is whirling in an anti-clockwise circle. Placed in the electron orbit, the atom creates a 30-degree angle in the magnetic field. Determine the circling electron’s torque.
Ans – As you know,
Magnetic moment related to electron M = eh/4πme & θ = 30°
τ = MBsinθ
is the final torque.
2) A charged particle traveling in a homogeneous magnetic field won’t change in energy. Justify!
Ans – The force acting on a charged element in a homogeneous magnetic field runs in a direction perpendicular to the direction of charge. Hence, the energy of the charged particle will not vary because the work performed by the magnetic field on the charge remains zero.
3) In a specific African region, a compass reads 12° west of geographic north. A dip circle kept in the magnetic meridian plane has its north-end magnetic needle denoting 55° above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Find the magnitude & direction of the earth’s field at the site.
Ans: As you know,
The angle of declination, θ = 12°
The angle of dip, δ = 60°
The horizontal component of earth’s magnetic field, BH = 0.16G
Earth’s magnetic field at the given location = B
We can equate B & BH as:
BH = Bcos δ
Earth’s magnetic field remains in the vertical plane, 12° west of the geographic meridian, creating an upward angle of 60° with the horizontal direction.
Therefore, its magnitude is denoted as 0.32G.
4) A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. The horizontal component of the earth’s magnetic field at the place is referred to as 0.35 G. Calculate the magnitude of the earth’s magnetic field.
Ans – As you know,
The horizontal component of earth’s magnetic field, BH = 0.35G
The angle created by the needle with the horizontal plane (angle of dip) = δ = 22°.
Earth’s magnetic field strength = B.
We can equate BH & B as,
BH = Bcos δ
⇒ B = BH/cos δ
⇒ B = 0.35/cos 22°
= 0.377G
Therefore, the strength of Earth’s magnetic field at a specific area is 0.377G.
5) What is the primary feature of the soft iron core implemented in a moving coil galvanometer? Given that a galvanometer offers full deflection for Ig. Is it possible to convert them into an ammeter of range I < Ig
Ans – A soft iron core is used in the moving coil galvanometer since it elevates the strength of the magnetic field by increasing the galvanometer sensitivity.
As you know,
For I < Ig, S becomes negative.
Hence, it cannot be changed into an ammeter of range I < Ig.
6) A short bar magnet of magnetic moment 0.9 J/T is kept with its axis at 60° to a uniform magnetic field. It experiences a torque of 0.063 Nm.
(i) Define the strength of the magnetic field
Ans – Since τ = MBsinθ (where θ= 60°, τ = 0.063Nm, M = 0.9J/T)
B = τ/(Msinθ) = 0.063/(0.9sin60°) = 0.081T
(ii) Calculate the bar magnet orientation related to the equilibrium position in the magnetic field.
Ans – The magnet remains in a stable equilibrium in the magnetic field when τ= 0.
i.e., MBsin0° = 0 => When the magnet aligns itself parallel to the field.
7) A short bar magnet of magnetic moment M = 0.32J/T is placed in a uniform magnetic field of 0.15T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its
a) Stable equilibrium: Calculate the potential energy of the magnet.
Ans – As you know,
The moment of the bar magnet, M = 0.32 J/T
External magnetic field, B = 0.15T
It is assumed to be in stable equilibrium when the bar magnet is aligned along the magnetic field.
Hence the angle θ, between the bar magnet & the magnetic field is 0°.
The potential energy of the system = -MBcosθ
⇒ – MBcosθ = – 0.32 x 0.15cos(0) = – 4.8 x 10 -2J
Therefore the potential energy will be – 4.8 x 10 -2J.
b) Unstable equilibrium: Calculate the potential energy of the magnet.
Ans: As you know,
The moment of the bar magnet, M = 0.32J/T
External magnetic field, B = 0.15T
When the bar magnet is aligned opposite to the magnetic field, it is considered to be in unstable equilibrium, θ = 180°.
Potential energy of the system is hence = -MBcosθ
⇒ – MBcosθ = – 0.32 x 0.15cos(180) = 4.8 x 10 -2J
Therefore the potential energy will be 4.8 x 10 -2J.
8) A lengthy straight horizontal cable has a current of 2.5A in the direction 10° south of west to 10° north of east. The magnetic meridian of the place happens to be 10º west of the geographic meridian. The earth’s magnetic field at the location is 0.33G with a zero-angle dip. Point the line of neutral points (Ignore the cable thickness, since at neutral points, the magnetic field due to a current-carrying cable is equal & opposite to the horizontal component of earth’s magnetic field.)
Ans – As you know,
The current in the wire, I = 2.5A
The angle of dip at the place, δ = 0°
The Earth’s magnetic field, H = 0.33G = 0.33 x 10-4 T
The horizontal component of Earth’s magnetic field is denoted as,
HH = Hcos δ = 0.33 x 10-4 x cos 0° = 0.33 * 10-4 T
The magnetic field at the neutral point at a distance R from the cable can be,
Where μ0 is the permeability of free space. So μ0 = 4π x 10-7Tm/A
= 15.15 x 10-3m = 1.51 cm
Hence, a series of neutral points remains on a straight line adjacent to the cable at a perpendicular distance of 1.51 cm over the plane of the paper.
9) A compass needle free to turn in a horizontal plane is placed at the centre of a circular coil of 30 turns & a radius of 12 cm. The coil is in a vertical plane creating a 45° angle with the magnetic meridian. Assume that the current in the coil is 0.35A & the needle directs from west to east.
a) Calculate the horizontal component of the Earth’s magnetic field.
Ans – As you know,
No. of turns in the circular coil given, N = 30
The radius of the circular coil given, r = 12 cm = 0.12 m
The current in the coil, I = 0.35A
The angle of dip, δ = 45°
The magnetic field due to the current I from a distance r can be denoted as:
Where μ0 is the permeability of free space. So μ0 = 4π x 10-7Tm/A
Therefore, B = 54.9 x 10-4T = 0.549 G
BH = B * sin δ
⇒ BH = 0.549 sin 45° = 0.388 G
The needle directs West to East. Therefore, the horizontal component of Earth’s magnetic field remains 0.388 G